# Tangent line of curve of intersection

## Homework Statement

Find the parametric equations of the line tangent to the curve of intersection of the paraboloid
z = x² + y² and the ellipsoid 4x² + y² + z² = 9 at the point ( -1, 1, 2 ).

## Homework Equations

Probable use of the gradient vector (as this is the chapter we are in)

## The Attempt at a Solution

I have found what I believe to be the curve of intersection by setting the equations of the surfaces equal to each other:

x² + y² - z = 4x² + y² + z² - 9
9 = 3x² + z² + z
F(x,y,z) = 3x² + z² + z - 9 (as a function)

< 6x, 0, 2z + 1 >

But I cannot figure out how to convert this from cartesian to polar (or if I even need to) and get the correct answer. I've tried several paths (i.e. substituting x=rcosσ) but my answer never matches the expected answer, which contains no trigonometry. Does the correct approach involve substituting x² + y² for z? This seems to be the most logical approach but I haven't been able to apply it correctly.

Filip Larsen
Gold Member
It is unclear to me why you want the tangent line in polar coordinates. The text seems only to call for a parametric line, eg. $\bar{r} = \bar{k} t +\bar{r}_0$.

Your right! I was thinking of a vector equation. Then I suppose the parametric equations would look something like: x = -1 +3t², y = 1, z = 3 + t² ? But that still doesn't contain the expected y component, I must be doing something wrong yet.

Filip Larsen
Gold Member
A line parametric in, say, t should be linear in t, otherwise its not a line. Think about which point the line passes through and what vector direction it "follows" from that point as t varies.

I remember why I wanted the vector equation. The only way I can think to find the tangent line is by taking the derivative of a vector equation.

Never mind. By crossing the gradients I found the tangent vector. This makes perfect sense, it was so simple :D :D